Is this operator a finite rank operator on $L^2$?

Question

Is the operator $T: L^2([0,1]) \rightarrow L^2([0,1])$ defined as: $$Tu(t) = t u(t)$$ a finite rank operator?

More in general: how can I understand if an operator $T$ on $L^2$ is a finite rank operator? Thanks!

Answer 1

Let $u_n(t):=t^{n-1}$ for $ n \in \mathbb N.$ Then $u_n \in L^2([0,1])$ for all $n$ and

$$Tu_n=u_{n+1}.$$ This shows that the functions $t,t^2,t^3,...$ are contained in $im(T).$

But $t,t^2,t^3,...$ are linearly independent in $L^2([0,1])$ , thus $T$ is not a finite rank operator.